Other time considerations

Source: Simon Garrett

Modifications by Evan Korth

2

Disk access time vs CPU time

• We have assumed that all instructions are created equally in our analysis all semester. What if this assumption is not true?

• One example:– It could be the case that our data does not fit in memory.

– In that case we need to access the secondary storage (disk)

3

A closer look at disk access

• Last year I purchased the components I need to build a home computer. My 750g drive spins at 7200RPM and my Intel Core2 Duo runs at 2.4 Ghz

• 7200 rpm means 7200 revolutions per minute or one revolution every 1/120 of a second (8.3ms)

• On average what we want is half way round this disk – it will take ~4ms.– Note I am assuming the read arm is in the correct position

• This sounds good until you realize that we get 240 disk accesses a second – the same time as 2.4 billion instructions (2.4 Ghz)

• In other words, one disk access takes about the same time as 10,000,000 instructions

• Not all instructions can be considered equal!

Note: processor speeds tend to improve at a quicker rate than Note: processor speeds tend to improve at a quicker rate than disk speeds so the cost of a disk access becomes more disk speeds so the cost of a disk access becomes more

expensive as related to CPU instructions over time.expensive as related to CPU instructions over time.

4

Solution

• Assume that we use an AVL tree to store all the car driver details in NYC (about 20 million records)

• Assume further that each node in the tree must be retrieved from the disk.

• We still end up with a very deep tree with lots of different disk accesses; log2 20,000,000 is about 24, so this takes about 0.1 seconds (if there is only one user of the program)

• We know we can’t improve on the log n for a binary tree

• But, the solution is to use more branches and thus less height!

• As branching increases, depth decreases

B-trees

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Definition of a B-tree

• A B-tree of order m is an m-way tree (i.e., a tree where each node may have up to m children) in which:1. the number of keys in each non-leaf node is one less than the number

of its children and these keys partition the keys in the children in the fashion of a search tree

2. all leaves are on the same level

3. all non-leaf nodes except the root have at least m / 2 children

4. the root is either a leaf node, or it has from two to m children

5. a leaf node contains no more than m – 1 keys

• The number m should always be odd

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An example B-Tree

51 6242

6 12

26

55 60 7064 9045

1 2 4 7 8 13 15 18 25

27 29 46 48 53

A B-tree of order 5 containing 26 items

Note that all the leaves are at the same levelNote that all the leaves are at the same level

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Inserting into a B-Tree

• Attempt to insert the new key into a leaf

• If this would result in that leaf becoming too big, split the leaf into two, promoting the middle key to the leaf’s parent

• If this would result in the parent becoming too big, split the parent into two, promoting the middle key

• This strategy might have to be repeated all the way to the top

• If necessary, the root is split in two and the middle key is promoted to a new root, making the tree one level higher

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• Suppose we start with an empty B-tree and keys arrive in the following order:1 12 8 2 25 5 14 28 17 7 52 16 48 68 3 26 29 53 55 45

• We want to construct a B-tree of order 5

• The first four items go into the root:

• To put the fifth item in the root would violate condition 5

• Therefore, when 25 arrives, pick the middle key to make a new root

Constructing a B-tree

1 2 8 12

10

Constructing a B-tree (contd.)

1 2

8

12 25

6, 14, 28 get added to the leaf nodes:

1 2

8

12 146 25 28

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Constructing a B-tree (contd.)

Adding 17 to the right leaf node would over-fill it, so we take the middle key, promote it (to the root) and split the leaf

8 17

12 14 25 281 2 6

7, 52, 16, 48 get added to the leaf nodes

8 17

12 14 25 281 2 6 16 48 527

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Constructing a B-tree (contd.)

Adding 68 causes us to split the right most leaf, promoting 48 to the root, and adding 3 causes us to split the left most leaf, promoting 3 to the root; 26, 29, 53, 55 then go into the leaves

3 8 17 48

52 53 55 6825 26 28 291 2 6 7 12 14 16

Adding 45 causes a split of 25 26 28 29

and promoting 28 to the root then causes the root to split

13

Constructing a B-tree (contd.)

17

3 8 28 48

1 2 6 7 12 14 16 52 53 55 6825 26 29 45

14

Inserting into a B-Tree

• Attempt to insert the new key into a leaf

• If this would result in that leaf becoming too big, split the leaf into two, promoting the middle key to the leaf’s parent

• If this would result in the parent becoming too big, split the parent into two, promoting the middle key

• This strategy might have to be repeated all the way to the top

• If necessary, the root is split in two and the middle key is promoted to a new root, making the tree one level higher

15

Exercise in Inserting a B-Tree

• Insert the following keys to a 5-way B-tree:

• 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56

• Check your approach with a neighbour and discuss any differences.

16

Removal from a B-tree

• During insertion, the key always goes into a leaf. For deletion we wish to remove from a leaf. There are three possible ways we can do this:

• 1 - If the key is already in a leaf node, and removing it doesn’t cause that leaf node to have too few keys, then simply remove the key to be deleted.

• 2 - If the key is not in a leaf then it is guaranteed (by the nature of a B-tree) that its predecessor or successor will be in a leaf -- in this case can we delete the key and promote the predecessor or successor key to the non-leaf deleted key’s position.

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Removal from a B-tree (2)

• If (1) or (2) lead to a leaf node containing less than the minimum number of keys then we have to look at the siblings immediately adjacent to the leaf in question: – 3: if one of them has more than the min’ number of keys then we can

promote one of its keys to the parent and take the parent key into our lacking leaf

– 4: if neither of them has more than the min’ number of keys then the lacking leaf and one of its neighbours can be combined with their shared parent (the opposite of promoting a key) and the new leaf will have the correct number of keys; if this step leaves the parent with too few keys then we repeat the process up to the root itself, if required

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Type #1: Simple leaf deletion

1212 2929 5252

22 77 99 1515 2222 5656 6969 72723131 4343

Delete 2: Since there are enoughkeys in the node, just delete it

Assuming a 5-wayB-Tree, as before...

Note when printed: this slide is animated

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Type #2: Simple non-leaf deletion

1212 2929 5252

77 99 1515 2222 5656 6969 72723131 4343

Delete 52

Borrow the predecessoror (in this case) successor

5656

Note when printed: this slide is animated

20

Type #4: Too few keys in node and its siblings

1212 2929 5656

77 99 1515 2222 6969 72723131 4343

Delete 72Too few keys!

Join back together

Note when printed: this slide is animated

21

Type #4: Too few keys in node and its siblings

1212 2929

77 99 1515 2222 696956563131 4343

Note when printed: this slide is animated

22

Type #3: Enough siblings

1212 2929

77 99 1515 2222 696956563131 4343

Delete 22

Demote root key andpromote leaf key

Note when printed: this slide is animated

23

Type #3: Enough siblings

1212

292977 99 1515

3131

696956564343

Note when printed: this slide is animated

24

Exercise in Removal from a B-Tree

• Given 5-way B-tree created by these data (last exercise):

• 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56

• Add these further keys: 2, 6,12

• Delete these keys: 4, 5, 7, 3, 14

• Again, check your approach with a neighbour and discuss any differences.

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Analysis of B-Trees

• The maximum number of items in a B-tree of order m and height h:root m – 1level 1 m(m – 1)level 2 m2(m – 1). . .level h mh(m – 1)

• So, the total number of items is

(1 + m + m2 + m3 + … + mh)(m – 1) =

[(mh+1 – 1)/ (m – 1)] (m – 1) = mmhh+1+1 – 1 – 1

• When m = 5 and h = 2 this gives 53 – 1 = 124

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Reasons for using B-Trees

• When searching tables held on disc, the cost of each disc transfer is high but doesn't depend much on the amount of data transferred, especially if consecutive items are transferred– If we use a B-tree of order 101, say, we can transfer each node in one

disc read operation

– A B-tree of order 101 and height 3 can hold 1014 – 1 items (approximately 100 million) and any item can be accessed with 3 disc reads (assuming we hold the root in memory)

• If we take m = 3, we get a 2-3 tree, in which non-leaf nodes have two or three children (i.e., one or two keys)– B-Trees are always balanced (since the leaves are all at the same

level), so 2-3 trees make a good type of balanced tree

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Comparing Trees

• Binary trees– Can become unbalanced and lose their good time complexity (big O)

– AVL trees are strict binary trees that overcome the balance problem

– Heaps remain balanced but only prioritise (not order) the keys

• Multi-way trees– B-Trees can be m-way, they can have any (odd) number of children

– One B-Tree, the 2-3 (or 3-way) B-Tree, approximates a permanently balanced binary tree, exchanging the AVL tree’s balancing operations for insertion and (more complex) deletion operations